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QUESTION

# How do I use DeMoivre's theorem to solve z^3-1=0?

If z^3-1=0, then we are looking for the cubic roots of unity, i.e. the numbers such that z^3=1.

If you're using complex numbers, then every polynomial equation of degree k yields exactly k solution. So, we're expecting to find three cubic roots.

De Moivre's theorem uses the fact that we can write any complex number as \rho e^{i \theta}= \rho (\cos(\theta)+i\sin(\theta)), and it states that, if z=\rho (\cos(\theta)+i\sin(\theta)), then z^n = \rho^n (\cos(n \theta)+i\sin(n \theta))

If you look at 1 as a complex number, then you have \rho=1, and \theta=2\pi. We are thus looking for three numbers such that \rho^3=1, and 3\theta=2\pi.

Since \rho is a real number, the only solution to \rho^3=1 is \rho=1. On the other hand, using the periodicity of the angles, we have that the three solutions for \theta are \theta_{1,2,3}=\frac{2k\pi}{3}, for k=0,1,2.

This means that the three solutions are:

1. \rho=1, \theta=0, which is the real number 1.
2. \rho=1, \theta=\frac{2\pi}{3}, which is the complex number -1/2 + \sqrt{3}/2 i
3. \rho=1, \theta=\frac{4\pi}{3}, which is the complex number -1/2 - \sqrt{3}/2 i